Lemma 33.25.11. Let $X$ be a scheme of finite type over a field $k$. There exists a finite purely inseparable extension $k'/k$, an integer $t \geq 0$, and closed subschemes

$X_{k'} \supset Z_0 \supset Z_1 \supset \ldots \supset Z_ t = \emptyset$

such that $Z_0 = (X_{k'})_{red}$ and $Z_ i \setminus Z_{i + 1}$ is smooth over $k'$ for all $i$.

Proof. We may use induction on $\dim (X)$. By Lemma 33.6.11 we can find a finite purely inseparable extension $k'/k$ such that $(X_{k'})_{red}$ is geometrically reduced over $k'$. By Lemma 33.25.7 there is a nowhere dense closed subscheme $X' \subset (X_{k'})_{red}$ such that $(X_{k'})_{red} \setminus X'$ is smooth over $k'$. Then $\dim (X') < \dim (X)$. By induction hypothesis there exists a finite purely inseparable extension $k''/k'$, an integer $t' \geq 0$, and closed subschemes

$X'_{k''} \supset Y_0 \supset Y_1 \supset \ldots \supset Y_{t'} = \emptyset$

such that $Y_0 = (X'_{k''})_{red}$ and $Y_ i \setminus Y_{i + 1}$ is smooth over $k''$ for all $i$. Then we let $t = t' + 1$ and we consider

$X_{k''} \supset Z_0 \supset Z_1 \supset \ldots \supset Z_ t = \emptyset$

given by $Z_0 = (X_{k''})_{red}$ and $Z_ i = Y_{i - 1}$ for $i > 0$; this makes sense as $X'_{k''}$ is a closed subscheme of $X_{k''}$. We omit the verification that all the stated properties hold. $\square$

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